# (Approximately) bijective proof of $\zeta(2)=\pi^2/6$ ?

Given $A,B\in {\Bbb Z}^2$, write $A \leftrightarrows B$ if the interior of the line segment AB misses ${\Bbb Z}^2$.

For $r>0$, define $S_r:=\{ \{A, B\} | A,B\in {\Bbb Z}^2,||A||<r,||B||<r, | ||A||-||B|| |<1 {\rm\ and\ } A \leftrightarrows B \}\ .$

A little calculus gives the equivalence of $\zeta(2)=\pi^2/6$ and $$\lim_{r\to\infty} \frac{|S_r|}{(2r)^3} = 1\ .$$

Of course $(2r)^3$ counts lattice points in a cube $C_{r}:=[-r,r)^3$.

Question: Does there exist an approximately bijective proof of this limit (or some variant), one that matches most of $S_r$ with most of $C_r$?

-
Do you have any reason to think there should be? –  Greg Martin May 2 '12 at 4:05
@Greg Martin 1) There do exist reasonably elementary evaluations of $\zeta(2)$ (no Fourier analysis, no contour integration, etc.) so perhaps it's possible to unwind one of them. 2) Any combinatorialist presented with a combinatorial identity naturally wonders whether a bijective proof exists - I don't know of any serious candidates for natural combinatorial facts with no natural proofs. –  David Feldman May 2 '12 at 5:46
Interesting question. Can you spell out the calculus a bit? I see various starts but not a clear path. –  Aaron Meyerowitz May 2 '12 at 7:11
Interesting question, does this generalize to $\zeta(2k)$? –  user22202 May 2 '12 at 7:16
Roughly (and not rigorously), each pair $A,B$ contains a first point reading clockwise, say $A$. The condition $|\,||A||−||B||\,|<1$ on $B$ in ${\Bbb R^2}$ locates $B$ in a half-annulus of area about $2\pi||A||$. This half-annulus contains a number of lattice points about equal to it's area, and of those, about $1/\zeta(2)$ satisfy $A⇆B$. Now do a double integral for the possible locations of $A$, first by norm, then by angle. The proportion of points where $A⇆B$ may not behave so will in a thin region, but some integration by parts makes everything legal. –  David Feldman May 2 '12 at 22:57
show 1 more comment